Hey folks!
I'm looking to prove that "eventually", $$n^{100}

Question

Hey folks!
I'm looking to prove that "eventually", $$n^{100} < n!$$ Conceptually, I can see that it is true once n gets large enough (about 125 according to wolfram alpha), since while individual terms on the denominator are less than or equal those of the numerator, the denominator will "eventually" have more terms than the numerator, which lets it grow bigger.

Any thoughts?

anonymous · Answer

I can just about see how that works for 100^k, but I'm not so clear on how it relates to k^100. Could you elaborate?

anonymous · Answer

I was thinking of using a rather fluffy argument, that uses proofs of simpler relations, like $$n < n!$$, and $$n^2 < n!$$ and simply using some kind of scale-up argument.

anonymous · Answer

i don't know if i am headed the right way but taking on each side sort of negated the inequality, since you have already mentioned it works for n>125, what are we supposed to do?
as in, its definitely not true for all n

anonymous · Answer

I know it's true for n>125 because Wolfram alpha told me it's true. I wanted to prove this fact, but in the end decided it was unnecessarily rigourous

anonymous · Answer

okay, i tried it taking logarithm on both sides and hit the wall because the proof depends on the value of n, thanks a lot